A067180 Smallest prime with digit sum n, or 0 if no such prime exists.

0, 2, 3, 13, 5, 0, 7, 17, 0, 19, 29, 0, 67, 59, 0, 79, 89, 0, 199, 389, 0, 499, 599, 0, 997, 1889, 0, 1999, 2999, 0, 4999, 6899, 0, 17989, 8999, 0, 29989, 39989, 0, 49999, 59999, 0, 79999, 98999, 0, 199999, 389999, 0, 598999, 599999, 0, 799999, 989999, 0, 2998999, 2999999, 0, 4999999

Offset: 1

Author

Amarnath Murthy, Jan 09 2002

Keywords

• easy
• nonn
• base

Examples

			a(68) = 59999999 because 59999999 is the smallest prime with digit sum = 68;
a(100) = 298999999999 because 298999999999 is the smallest prime with digit sum = 100.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..1000 (first 175 terms from Robert G. Wilson v)

Crossrefs

Cf. A054750.

Removal of the 0 terms from this sequence leaves A067523.

Programs

Maple

g:= proc(s,d) # integers of <=d digits with sum s
  if s > 9*d then return [] fi;
  if d = 1 then return [s] fi;
  [seq(op(map(t -> j*10^(d-1)+ t, g(s-j,d-1))),j=0..9)];
end proc:
f:= proc(n) local d, j,x,y;
  if n mod 3 = 0 then return 0 fi;
  for d from ceil(n/9) do
    if d = 1 then
      if isprime(n) and n < 10 then return n
      else next
      fi
    fi;
    for j from 1 to 9 do
      for y in g(n-j,d-1) do
        x:= 10^(d-1)*j + y;
        if isprime(x) then return x fi;
  od od od;
end proc:
f(1):= 0: f(3):= 3:
map(f, [$1..100]); # Robert Israel, Dec 13 2020

Mathematica

a = Table[0, {100}]; Do[b = Apply[ Plus, IntegerDigits[ Prime[n]]]; If[b < 101 && a[[b]] == 0, a[[b]] = Prime[n]], {n, 1, 10^7} ]; a
f[n_] :=  If[n > 5 && Mod[n, 3] == 0, 0, Block[{k = 1, lmt, lst = {}, ip = IntegerPartitions[n, Round[1 + n/9], {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}]}, lmt = 1 + Length@ ip; While[k < lmt, AppendTo[lst, Select[ FromDigits@# & /@ Permutations@ ip[[k]], PrimeQ[#] &]]; k++]; Min@ Flatten@ lst]]; f[1] = 0; f[4] = 13; Array[f, 70] (* Robert G. Wilson v, Sep 28 2014 *)

PARI

A067180(n)={if(n<2, 0, n<4, n, n%3, my(d=divrem(n,9)); forprime(p=d[2]*10^d[1]-1,,sumdigits(p)==n&&return(p)))} \\ M. F. Hasler, Nov 04 2018

Formula

a(3k) = 0 for k > 1.

a(3k-2) = A067523(2k-1), a(3k-1) = A067523(2k), for all k > 1. - M. F. Hasler, Nov 04 2018

Extensions

Edited and extended by Robert G. Wilson v, Mar 01 2002

Edited by Ray Chandler, Apr 24 2007